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Liquidammonia is used in ice factory for making ice from water. If water at20^(@)Cis to be converted into 2 kg ice at 0^(@)C , how many grams of ammonia are to be evaporated ? (Given : The latent heat of vaporization of ammonia = 341 cal //g) |
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Answer» Solution :Data : `m_(1) = 2 kg , DeltaT_(1) = 20^(@) C -0^(@) C = 20^(@)C` ` c_(1) = 1" kcal/kg."^(@)C , L_(1) ("ice") = 80 "kcal/kg", L_(2)` (vaporization of ammonia) = 341 cal/g = 341 kcal/kg,`m_(2) = ? ` `Q_(1)` = (heat lost by water) = `m_(1)c_(1)DeltaT_(1) + m_(1)L_(1)` 2 kg `xx 1 kcal//kg.^(@)C xx 20^(@)C + 2kg xx 80` kcal/kg 40 Kcal + 160 kcal = 200 kcal ` Q_(2)` ( heat absorbed by ammonia) = `m_(2)L_(2) =m_(2) xx 341` kcal/ kg ltbRgt According to the principal of heatexchange , ` Q_(1) = Q_(2)` 200 kcal. `= m_(2) xx 341` kcal/ kg `m_(2) = 200/341 " kg" =` 0.5864 kg =586 .4 g 586.4 g of ammonia are to be evaporated . |
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